|
|
var a=[1,2,3,4,5,6];
- s/ e, Z* U4 K! {" A9 l! j' evar b=[2,3,6];7 {! `, \( b9 {" }0 G" Z/ R, O
var c =[];! G: Y6 T6 w8 n
for(var i=0;i<a.length;i++){0 ?# c' x* k( |1 r T
var has = false;+ ]. [0 t. |# Q9 @/ _. g; P
var data = a[i];/ `/ c. b2 }6 d2 Y3 t
//判断baic中是否有du元素data
# I" ^* z" l# W* a* Lfor(var j=0;j<c.length;j++){
0 _; J) t. N/ Q7 ?; Yif(data ==c[j]){
/ Z: U! ?9 r+ G- qhas =true;/ n, Z7 C Z) v" V3 g+ G
break;
( A0 ?$ O; M7 y- Z. B+ a}$ j+ y2 S" r5 Q& l# z, K$ f
}
- ^' _3 w+ j p5 j2 f! g//如果zhi没有5 U9 _: E: J( L% R. y c- [
if(has==false){9 `" a: V' b* H( q' V
c.push(data);
- U+ q) e! s$ w, V* V( T/ I: [}- [% k9 |3 @" p" i
}
1 ^* z" K* \4 b1 A* ~for(var i=0;i<b.length;i++){! M" v. V" c$ ~ @3 r4 V
bool has = false;; g2 v) E5 x \) L) X" I
var data = a[i];" L6 b3 B/ F3 d# \. l
//判断c中是否有元素data
! u' c/ P2 b/ c; K5 }; [for(var j=0;j<b.length;j++){! Z3 z- n+ ]3 R, M% U7 k& c
if(data ==b[j]){1 o& s# P/ G2 y8 C
has =true;
: I y; x `8 c) M- h, [break;+ X7 j4 C9 J, s2 }+ r; S: X
}+ [2 V- A/ b8 G% f0 M
}
: a. t/ P2 h+ w( u9 O' A//如果没有
/ l- M/ E. |8 K) V* |1 P/ i C7 H- aif(has==false){/ M M# t6 i3 l% b/ X2 @
c.push(data);
0 m2 M) E1 z7 t}
* }& |: T- r/ d, {% |}
. w( b9 ]5 I/ y//最后c就是结果 |
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/1 
发表于 2020-7-6 13:49
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淘帖
支持！
反对！